IGCSE Mathematics 0580 Paper 4 Practice Paper (8 Questions, 65 Marks, 90 Minutes)

2026-11-02

IGCSE Mathematics 0580 Paper 4

IGCSE Mathematics 0580 Paper 4 practice paper: 8 original questions in 90 minutes, covering simultaneous and quadratic equations, factorisation, the sine and cosine rules.

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Paper details　Level: IGCSE (Grade 9-10) | Subject: Mathematics | Time: 90 minutes | Marks: 65 | Questions: 8

Section A Algebra, trigonometry and statistics (30 marks)

Questions 1-4 (algebra, trigonometry and statistics; marks shown per part)

#	Question	Answer and marking notes
1	Solve the simultaneous equations 2x + 3y = 19 and 4x – y = 17. (a) Solve the equations by elimination. (3 marks) (b) Hence find the value of 6x + 2y. (2 marks) (c) Solve the quadratic equation x² – 7x + 12 = 0. (2 marks)	(a) From 4x – y = 17, y = 4x – 17 (1 mark); substituting into 2x + 3(4x – 17) = 19 gives 14x = 70, so x = 5 (1 mark); y = 4(5) – 17 = 3 (1 mark). (b) 6x + 2y = 6(5) + 2(3) = 36 (1 mark substitution, 1 mark answer). (c) (x – 3)(x – 4) = 0, so x = 3 or x = 4 (1 mark factorising, 1 mark both roots).
2	Algebraic manipulation. (a) Expand (2x – 3)(x + 5). (2 marks) (b) Factorise 6x² – 11x – 10. (2 marks) (c) Simplify (x² – 9) / (x² + x – 12). (2 marks) (d) Solve 3 / (x – 1) = 2 / (x + 2). (2 marks)	(a) 2x² + 10x – 3x – 15 = 2x² + 7x – 15 (1 mark expansion, 1 mark simplified). (b) 6x² – 15x + 4x – 10 = (3x + 2)(2x – 5) (1 mark grouping, 1 mark answer). (c) (x – 3)(x + 3) / [(x + 4)(x – 3)] = (x + 3) / (x + 4) (1 mark factorising, 1 mark simplified; state x ≠ 3 and x ≠ -4). (d) 3(x + 2) = 2(x – 1), so 3x + 6 = 2x – 2 and x = -8 (1 mark cross-multiplying, 1 mark answer).
3	A ladder of length 6.5 m leans against a vertical wall and makes an angle of 68° with the horizontal ground. (a) Find the height of the top of the ladder above the ground, correct to 0.01 m. (2 marks) (b) Find the distance from the foot of the ladder to the wall, correct to 0.01 m. (2 marks) In triangle ABC, AB = 9.6 cm, AC = 6.9 cm and angle BAC = 52°. (c) Find the length of BC, correct to 0.01 cm. (2 marks) (d) Use the sine rule to find angle ABC, correct to 0.1°. (2 marks)	(a) Height = 6.5 × sin 68° = 6.03 m (1 mark correct ratio, 1 mark answer). (b) Distance = 6.5 × cos 68° = 2.43 m (1 mark correct ratio, 1 mark answer). (c) BC² = 9.6² + 6.9² – 2 × 9.6 × 6.9 × cos 52° = 58.21, so BC = 7.63 cm (1 mark cosine rule, 1 mark answer). (d) sin B / 6.9 = sin 52° / 7.63, so sin B = 0.713 and angle ABC = 45.5° (1 mark sine rule, 1 mark answer).
4	The daily study times, in minutes, of 15 students are: 12, 14, 15, 17, 18, 19, 21, 23, 24, 26, 28, 30, 33, 35, 38. (a) Calculate the mean, correct to 0.1 minutes. (2 marks) (b) State the median. (1 mark) (c) Calculate the interquartile range. (3 marks) (d) Compare the mean with the median and state, with a reason, whether the distribution is positively skewed. (1 mark)	(a) Mean = 353 / 15 = 23.5 minutes (1 mark total, 1 mark answer). (b) With 15 values the median is the 8th value, 23 minutes (1 mark). (c) The lower quartile is the (15 + 1) / 4 = 4th value, 17, and the upper quartile is the 12th value, 30, so the interquartile range is 30 – 17 = 13 minutes (1 mark positions, 1 mark both quartiles, 1 mark difference). (d) The mean, 23.5, is greater than the median, 23, because a few large values pull it upwards, so the distribution is positively skewed (1 mark).

Section B Geometry, mensuration, percentages and probability (35 marks)

Questions 5-8 (geometry, mensuration, percentages and probability; marks shown per part)

#	Question	Answer and marking notes
5	A, B, C and D are points on a circle in that order and angle ABC = 118°. (a) Find angle ADC. (2 marks) (b) O is the centre of the circle; find the smaller angle AOC. (2 marks) AT is the tangent to the circle at A, and T and D lie on opposite sides of chord AC, with angle TAC = 62°. (c) State the relationship between angle TAC and angle ADC, naming the theorem used. (2 marks) (d) Find angle OAC. (3 marks)	(a) ABCD is a cyclic quadrilateral, so opposite angles are supplementary: angle ADC = 180° – 118° = 62° (1 mark theorem, 1 mark answer). (b) The angle at the centre is twice the angle at the circumference on the same arc AC: angle AOC = 2 × 62° = 124° (1 mark relationship, 1 mark answer). (c) By the alternate segment theorem the angle between the tangent AT and chord AC equals the angle in the alternate segment, so angle TAC = angle ADC = 62° (1 mark named theorem, 1 mark comparison). (d) A radius is perpendicular to the tangent at the point of contact, so angle OAT = 90° (1 mark); angle OAC = 90° – 62° = 28° (1 mark subtraction, 1 mark answer; check 2 × 28° + 124° = 180°).
6	A solid is made from a cylinder of radius 7 cm and height 20 cm with a cone of the same radius and slant height 25 cm on top. (a) Find the height of the cone. (2 marks) (b) Find the total volume of the solid, in terms of pi. (3 marks) (c) Find the curved surface area of the solid, excluding the base, in terms of pi. (2 marks) (d) The whole solid is melted down and recast as spheres of radius 7 cm. What is the greatest number of complete spheres that can be made? (2 marks)	(a) Height = √(25² – 7²) = √576 = 24 cm (1 mark Pythagoras, 1 mark answer). (b) Cylinder volume = pi × 7² × 20 = 980pi; cone volume = one third × pi × 7² × 24 = 392pi; total = 1372pi cubic cm (about 4310 cubic cm) (1 mark cylinder, 1 mark cone, 1 mark sum). (c) Curved surface area = 2 × pi × 7 × 20 + pi × 7 × 25 = 455pi square cm (1 mark both parts, 1 mark answer). (d) Sphere volume = four thirds × pi × 7³ = 457pi; 1372pi ÷ 457.33pi = 3, so 3 complete spheres can be made (1 mark sphere volume, 1 mark division).
7	(a) An item is priced at $480. Its price is reduced by 25% and then the reduced price is increased by 15%. Find the final selling price. (2 marks) (b) Find the percentage change from the original price to the final selling price. (2 marks) (c) $5000 is invested at 3.2% per year compound interest, compounded annually. Find the amount after 4 years, correct to the nearest dollar. (2 marks) (d) After how many whole years does the amount first exceed $6,000? (2 marks)	(a) 480 × (1 – 25%) = 360, then 360 × (1 + 15%) = 413 dollars (1 mark reduction, 1 mark increase). (b) (413 – 480) / 480 × 100% = -13.75% (1 mark substitution, 1 mark answer; the negative sign shows a net decrease). (c) Amount = 5000 × 1.032⁴ = 5671.38, about $5671 (1 mark formula, 1 mark answer). (d) After 5 years the amount is $5853, which is below 6,000; after 6 years it is $6040, the first year above 6,000, so 6 years are needed (1 mark checking years, 1 mark answer).
8	In a school of 40 students, 25 study Mathematics, 18 study Physics and 7 study both subjects. (a) Find the number who study Mathematics only, Physics only, and neither subject. (3 marks) (b) A student is chosen at random; find the probability that the student studies exactly one of the two subjects. (2 marks) (c) Given that the chosen student studies Mathematics, find the probability that the student also studies Physics. (2 marks) (d) Solve y = 2x – 1 and y = x² – 4 simultaneously for the x-coordinates of the points of intersection. (2 marks)	(a) Mathematics only = 25 – 7 = 18; Physics only = 18 – 7 = 11; at least one subject = 36, so neither = 40 – 36 = 4 (1 mark each). (b) Exactly one subject covers 29 students, so the probability is 29 / 40 = 0.725 (1 mark numerator, 1 mark answer). (c) Of the 25 Mathematics students, 7 also study Physics, so the probability is 7 / 25 = 0.28 (1 mark conditional probability, 1 mark answer). (d) x² – 4 = 2x – 1 gives x² – 2x – 3 = 0, so (x – 3)(x + 1) = 0 and x = 3 or x = -1 (1 mark equation, 1 mark both solutions).

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